Question

A particle starts from rest and moving with an acceleration of $4{\text{ms}}^{–2}$ and covers a distance of $15\text{m}$. The final velocity of the particle will be

• A. $\sqrt{120}{\text{ms}}^{-1}$
• B. $\sqrt{240}{\text{ms}}^{-1}$
• C. $\sqrt{360}{\text{ms}}^{-1}$
• D. $\sqrt{180}{\text{ms}}^{-1}$

Answer 1

The correct option is A $\sqrt{120}{\text{ms}}^{-1}$

Given,
$u=0,v=?,a=4{\text{ms}}^{-2},s=15\text{m}$
We know that,
$v^2-u^2=2as$
$\therefore v^2-0=2\times 4\times 15$
$v^2=120$
$v=\sqrt{120}{\text{ms}}^{-1}$